A conveyor belt is moving at a constant speed of `2m//s` . A box is grenty dropped on it. Th ecoefficient of friction between them is `mu=0.5` . The distance that the box will move relative to belt before coming to rest on it taking `g=10ms^(-2)` is:

Options

A. `1.2m`
B. `0.6m`
C. zero
D. `0.4m`

Explanation

To solve the problem of how far the box will move relative to the conveyor belt before coming to rest, we can follow these steps:

### Step 1: Identify the initial conditions
The conveyor belt is moving at a constant speed of \( v_{c} = 2 \, \text{m/s} \). When the box is dropped onto the conveyor, it also has an initial velocity of \( v_{0} = 2 \, \text{m/s} \) (the same as the conveyor).

### Step 2: Determine the forces acting on the box
The only horizontal force acting on the box after it is dropped is the frictional force, which will act in the opposite direction to the motion of the box relative to the conveyor. The frictional force \( F_f \) can be calculated using the formula:
\[ F_f = \mu \cdot N \]
where \( \mu = 0.5 \) is the coefficient of friction and \( N \) is the normal force. The normal force \( N \) is equal to the weight of the box, \( mg \).

### Step 3: Calculate the frictional force
Since \( N = mg \), we have:
\[ F_f = \mu \cdot mg \]
Substituting the values:
\[ F_f = 0.5 \cdot mg \]

### Step 4: Determine the acceleration due to friction
The acceleration \( a \) caused by the frictional force can be found using Newton's second law:
\[ F = ma \implies a = \frac{F_f}{m} \]
Substituting for \( F_f \):
\[ a = \frac{0.5 \cdot mg}{m} = 0.5g \]
Given \( g = 10 \, \text{m/s}^2 \):
\[ a = 0.5 \cdot 10 = 5 \, \text{m/s}^2 \]
This acceleration acts in the opposite direction to the motion of the box, so we can consider it as a deceleration.

### Step 5: Use the kinematic equation to find the distance
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \, \text{m/s} \) (final velocity when the box comes to rest),
- \( u = 2 \, \text{m/s} \) (initial velocity of the box),
- \( a = -5 \, \text{m/s}^2 \) (deceleration due to friction).

Rearranging the equation to solve for \( s \):
\[ 0 = (2)^2 + 2(-5)s \]
\[ 0 = 4 - 10s \]
\[ 10s = 4 \]
\[ s = \frac{4}{10} = 0.4 \, \text{m} \]

### Final Answer
The distance that the box will move relative to the belt before coming to rest is \( 0.4 \, \text{m} \).

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